7 Let the Titrations Begin

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TITRATION ON MARS

Robotic arm of Phoenix Mars Lander scoops up soil for chemical analysis on Mars. [NASA/JPL- Caltech/University of Arizona/Texas A&M University.]

In 2008, Professor Sam Kounaves and his students at Tufts University had the thrill of a life- time as their Wet Chemistry Laboratory aboard the Phoenix Mars Lander returned a stream of information about the ionic composition of Martian soil scooped up by a robotic arm. The arm delivered ~1 gram of soil through a sieve into a “beaker” fi tted with a suite of electro- chemical sensors described in Chapter 15. Aqueous solution added to the beaker shown in Box 15-3 leached soluble salts from the soil, while sensors measured ions appearing in the liquid. Unlike other ions, sulfate was measured by a precipitation titration with Ba21: S 21 2 FreemanBaCl2(s) Ba 1 2Cl 22 21 S SO4 1 Ba BaSO4(s)

As solid BaCl2 from a reagent canister slowly dissolved in the aqueous liquid, BaSO4 precipitated. In Problem 7-21, we see that one sensor showed a low level of Ba21 until 22 enough reagent had been added to react with all SO4 . Another sensor found steadily increas- 2 ing Cl as BaCl2 dissolved. The end point of the titration is marked by a sudden increase of 21 22 Ba when the last SO4 has precipitated and BaCl2 continues to dissolve. The increase in 2 Cl from the beginning of the titration up to the end point tells how much BaCl2 was required 22 to consume SO4 . Titrations of two soil samples in two cells found ,1.3 (60.5) wt% sulfate 1 W.H.in the soil. Other evidence suggests that the sulfate is mostly from MgSO4.

rocedures in which we measure the volume of reagent needed to react with analyte are Pcalled volumetric analysis. In this chapter, we discuss principles that apply to all volumetric procedures and then focus on precipitation titrations. Acid-base, oxidation-reduction, complex formation, and spectrophotometric titrations are discussed later in the book in their respective chapters.

7-1 Titrations In a titration, increments of reagent solution—the titrant—are added to analyte until their reaction is complete. From the quantity of titrant required, we can calculate the quantity of analyte that must have been present. Titrant is usually delivered from a buret (Figure 7-1).

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The principal requirements for a titration reaction are that it have a large equilibrium constant and proceed rapidly. That is, each increment of titrant should be completely and quickly consumed by analyte until the analyte is used up. Common titrations rely on acid- base, oxidation-reduction, complex formation, or precipitation reactions. The equivalence point occurs when the quantity of added titrant is the exact amount necessary for stoichiometric reaction with the analyte. For example, 5 mol of oxalic acid react with 2 mol of permanganate in hot acidic solution:

O O ™ ™ ® ® ® B 2 Reaction 7-1 is an oxidation-reduction reaction. 5HO C C OH 2MnO4 6H 10CO2 2Mn 8H2O (7-1) Please study Appendix D if you need to relearn how to balance Reaction 7-1. Analyte Titrant Oxalic acid Permanganate colorless colorless colorless purple

If the unknown contains 5.000 mmol of oxalic acid, the equivalence point is reached when 2 2.000 mmol of MnO 4 have been added. The equivalence point is the ideal (theoretical) result we seek in a titration. What we actu- ally measure is the end point, which is marked by a sudden change in a physical property of the solution. In Reaction 7-1, a convenient end point is the abrupt appearance of the purple color of permanganate in the fl ask. Prior to the equivalence point, all permanganate is con- Level of titrant sumed by oxalic acid, and the titration solution remains colorless. After the equivalence point, 2 unreacted MnO4 accumulates until there is enough to see. The fi rst trace of purple color is the end point. The better your eyes, the closer will be your measured end point to the true equivalence point. Here, the end point cannot exactly equal the equivalence point, because extra Buret 2 clamp MnO4, beyond that needed to react with oxalic acid, is required to exhibit purple color. Methods for determining when the analyte has been consumed include (1) detecting a Buret sudden change in the voltage or current between a pair of electrodes (Figure 7-5), (2) observing an indicator color change (Color Plate 2), and (3) monitoring absorption of light (Figure 18-11). An indicator is a compound with a physical property (usually color) that changes abruptly near the equivalence point. The change is caused by the disappearance of analyte or the appearance of excess titrant. The difference between the end point and the equivalence point is an inescapable titration error. By choosing a physical property whose change is easily observed (such as pH or Stopcock the color of an indicator), the end point can be very close to the equivalence point. We estimate the titration error with a blank titration, in which we carry out the same procedure without analyte. For example,Freeman we can titrate a solution containing no oxalic acid to see how 2 much MnO 4 is needed to produce observable purple color. We then subtract this volume of 2 MnO 4 from the volume observed in the analytical titration. The validity of an analytical result depends on knowing the amount of one of the reactants used. If a titrant is prepared by dissolving a weighed amount of pure reagent in a known Flask volume of solution, its concentration can be calculated. We call such a reagent a primary standard, because it is pure enough to be weighed and used directly. A primary standard should be 99.9% pure, or better. It should not decompose under ordinary storage, and it should be stable when dried by heat or vacuum, because drying is required to remove traces Solution of water adsorbed from the atmosphere. The highest quality, commercially available, primary of analyte Magnetic stirring standard pure materials, such as As2O3 (arsenic(III) oxide), CaCO3, Hg metal, and Ni metal, W.H.bar have purities certifi ed to within 60.05%. Primary standards for many elements are given in Appendix K. Box 7-1 discusses reagent purity. Box 3-2 describes certifi ed reference materials that allow laboratories to test the accuracy of their procedures. Many reagents used as titrants, such as HCl, are not available as primary standards. Instead, we prepare titrant with approximately the desired concentration and use it to titrate a primary standard. By this procedure, called standardization, we determine the concentration Magnetic of titrant. We then say that the titrant is a standard solution. The validity of the analytical stirrer result ultimately depends on knowing the composition of a primary standard. Sodium oxalate (Na C O ) is a commercially available primary standard for generating oxalic acid to stan- FIGURE 7-1 Typical setup for a titration. The 2 2 4 analyte is contained in the ﬂ ask, and the titrant dardize permanganate in Reaction 7-1. is in the buret. The stirring bar is a magnet In a direct titration, we add titrant to analyte until the reaction is complete. Occasion- coated with Teﬂ on, which is inert to most ally, we perform a back titration, in which we add a known excess of one standard reagent solutions. The bar is spun by a rotating magnet to the analyte. Then we titrate the excess reagent with a second standard reagent. A back inside the stirring motor. titration is useful when its end point is clearer than the end point of the direct titration, or

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BOX 7-1 Reagent Chemicals and Primary Standards Chemicals are sold in many grades of purity. For analytical analytical chemistry, carry designations such as “chemically pure” chemi stry, we usually use reagent-grade chemicals meeting (CP), “practical,” “purifi ed,” or “technical.” purity requirements set by the American Chemical Society (ACS) A few chemicals are sold in high enough purity to be primary Committee on Analytical Reagents.2 Sometimes “reagent grade” standard grade. Whereas reagent-grade potassium dichromate has a simply meets purity standards set by the manufacturer. An actual lot lot assay of $99.0%, primary standard grade K2Cr2O7 must be in analysis for specifi ed impurities should appear on the reagent bottle. the range 99.95–100.05%. Besides high purity, a key quality of For example, here is a lot analysis of zinc sulfate: primary standards is that they are indefi nitely stable. For trace analysis (analysis of species at ppm and lower levels), impurities in reagent chemicals must be extremely low. For this ZnSO4 ACS Reagent Lot Analysis: purpose, we use very-high-purity, expensive grades of acids such as Assay: 100.6% Fe: 0.000 5% Ca: 0.001% “trace metal grade” HNO3 or HCl to dissolve samples. We must pay Insoluble matter: careful attention to reagents and vessels whose impurity levels could 0.002% Pb: 0.002 8% Mg: 0.000 3% be greater than the quantity of analyte we seek to measure. pH of 5% solution The following steps help you to protect the purity of chemical at 25°C: 5.6 Mn: 0.6 ppm K: 0.002% reagents: Ammonium: • Avoid putting a spatula into a bottle. Instead, pour chemical out 0.000 8% Nitrate: 0.000 4% Na: 0.003% of the bottle into a clean container (or onto weighing paper) and Chloride: 1.5 ppm dispense the chemical from the clean container. • Never pour unused chemical back into the reagent bottle. • Replace the cap on the bottle immediately to keep dust out. The assay value of 100.6% means that a specifi ed analysis for one • Never put a glass stopper from a liquid-reagent container down of the major components produced 100.6% of the theoretical on the lab bench. Either hold the stopper or place it in a clean value. For example, if ZnSO4 is contaminated with the lower- place (such as a clean beaker) while you dispense reagent. 21 molecular-mass Zn(OH)2, the assay for Zn will be higher than the • Store chemicals in a cool, dark place. Do not expose them value for pure ZnSO4. Less pure chemicals, generally unsuitable for unnecessarily to sunlight.

when an excess of the fi rst reagent is required for complete reaction with analyte. To appreci- ate the difference between direct and back titrations, consider fi rst the addition of permanganate titrant to oxalic acid analyte in Reaction 7-1; this reaction is a direct titration. Alternatively, to perform a back titration, we could add a known excess of permanganate to consume oxalic acid. Then we could back titrate the excess permanganate with standard Fe21 to measure how much permanganate was left after reactionFreeman with the oxalic acid. An alternative to measuring titrant by volume is to measure the mass of titrant solution delivered in each increment. We call this procedure a gravimetric titration. Titrant can be delivered from a pipet. Titrant concentration is expressed as moles of reagent per kilogram of solution. Precision is improved from 0.3% attainable with a buret to 0.1% with a balance (see the dilution example on page 55). Experiments by Guenther and by Butler and Swift provide examples.3 “Gravimetric titrations should become the gold standard, and volumetric glass- ware should be seen in museums only.” 4

7-2 Titration Calculations Here are some examples to illustrate stoichiometry calculations in volumetric analysis. The key step is to relate molesW.H. of titrant to moles of analyte.

EXAMPLE Standardization of Titrant Followed by Analysis of Unknown The calcium content of urine can be determined by the following procedure: Step 1 Precipitate Ca21 with oxalate in basic solution: 21 22 S Ca 1 C2O4 Ca(C2O4) ? H2O(s) Oxalate Calcium oxalate Step 2 Wash the precipitate with ice-cold water to remove free oxalate, and dissolve the 21 solid in acid to obtain Ca and H2C2O4 in solution. Step 3 Heat the solution to 608C and titrate the oxalate with standardized potassium permanganate until the purple end point of Reaction 7-1 is observed.

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Standardization Suppose that 0.356 2 g of Na2C2O4 is dissolved in a 250.0-mL volumetric fl ask. If 10.00 mL of this solution require 48.36 mL of KMnO4 solution for titration, what is the molarity of the permanganate solution?

Solution The concentration of the oxalate solution is

0.356 2 g Na2C2O4 /(134.00 g Na2C2O4 /mol) We show an extra, subscripted digit for 5 0.010 633 M calculations. In general, retain all extra digits 0.250 0 L in your calculator. Do not round off until the 22 24 The moles of C2O4 in 10.00 mL are (0.010 633 mol/L)(0.010 00 L) 5 1.0633 3 10 mol 5 end of a problem. 0.106 33 mmol. Reaction 7-1 requires 2 mol of permanganate for 5 mol of oxalate, so the 2 MnO4 delivered must have been 2 mol MnO2 2 2 4 22 Reaction 7-1 requires 2 mol MnO 4 for 5 mol Moles of MnO 4 5 22 (mol C2O4 ) 5 0.042 531 mmol 22 a5 mol C O b C2O4 2 4 2 The concentration of MnO 4 in the titrant is therefore

2 0.042 531 mmol 2 mmol mol Molarity of MnO 5 5 8.794 3 10 4 M Note that is the same as . 4 7 mL L 48.36 mL 22 Analysis of Unknown Calcium in a 5.00-mL urine sample was precipitated with C2O4 2 and redissolved; the sample then required 16.17 mL of standard MnO 4 solution. Find the concentration of Ca21 in the urine.

2 24 Solution In 16.17 mL of MnO4, there are (0.016 17 L)(8.7947 3 10 mol/L) 5 1.4221 3 25 2 10 mol MnO4. This quantity will react with 5 mol C O22 22 22 2 4 2 Reaction 7-1 requires 5 mol C2O4 for 2 mol Moles of C2O4 5 2 (mol MnO4 ) 5 0.035 553 mmol 2 a 2 mol MnO4 b MnO 4

Because there is one oxalate ion for each calcium ion in Ca(C2O4) ?H2O, there must have 21 been 0.035 553 mmol of Ca in 5.00 mL of urine: 0.035 55 mmol [Ca21] 5 3 5 0.007 11 M 5.00 mL 1

TEST YOURSELF In standardization, 10.00 mL of Na2C2O4 solution required 39.17 mL of 2 KMnO4. Find the molarity of KMnO4. The unknown required 14.44 mL of MnO 4. Find [Ca21] in the urine.Freeman (Answer: 1.086 3 1023 M, 7.840 3 1023 M)

EXAMPLE Titration of a Mixture Solving for two unknowns requires two A solid mixture weighing 1.372 g containing only sodium carbonate and sodium bicarbon- independent pieces of information. Here we ate required 29.11 mL of 0.734 4 M HCl for complete titration: have the mass of the mixture and the volume S of titrant. Na2CO3 1 2HCl 2NaCl(aq) 1 H2O 1 CO2 FM 105.99 S NaHCO3 1 HCl NaCl(aq) 1 H2O 1 CO2 W.H. FM 84.01 Find the mass of each component of the mixture.

Solution Let’s denote the grams of Na2CO3 by x and grams of NaHCO3 by 1.372 2 x. The moles of each component must be x g (1.372 2 x) g Moles of Na CO 5 Moles of NaHCO 5 2 3 105.99 g/mol 3 84.01 g/mol We know that the total number of moles of HCl used was (0.029 11 L)(0.734 4 M) 5 0.021 38 mol. From the stoichiometry of the two reactions, we can say that

2(mo l Na2CO3) 1 mol NaHCO3 5 0.021 38 x 1.372 2 x 2 1 5 0.021 38 1 x 5 0.724 g a105.99b 84.01

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The mixture contains 0.724 g of Na2CO3 and 1.372 2 0.724 5 0.648 g of NaHCO3.

TEST YOURSELF A 2.000-g mixture containing only K2CO3 (FM 138.21) and KHCO3 (FM 100.12) required 15.00 mL of 1.000 M HCl for complete titration. Find the mass of each component of the mixture. (Answer: K2CO3 5 1.811 g, KHCO3 5 0.189 g)

7-3 Precipitation Titration Curves 2 In gravimetric analysis, we could measure an unknown concentration of I by adding excess Please review Section 6-3 on the solubility Ag1 and weighing the AgI precipitate [I2 1 Ag1 S AgI(s)]. In a precipitation titration, we product prior to studying precipitation titration monitor the course of the reaction between analyte (I2) and titrant (Ag1) to locate the equiva- curves. lence point at which there is exactly enough titrant for stoichiometric reaction with the analyte. Equivalence point: Point at which Knowing how much titrant was added tells us how much analyte was present. We seek the stoichiometric amounts of reactants have equivalence point in a titration, but we observe the end point at which there is an abrupt change been mixed. in a physical property (such as an electrode potential) that is being measured. The physical End point: Point near the equivalence point property is chosen to make the end point as close as possible to the equivalence point. at which an abrupt change in a physical The titration curve is a graph showing how the concentration of a reactant varies as property is observed. titrant is added. We will derive equations that can be used to predict precipitation titration curves. One reason to calculate titration curves is to understand the chemistry that occurs during titrations. A second reason is to learn how experimental control can be exerted to infl uence the quality of an analytical titration. Concentrations of analyte and titrant and the size of the solubility product (Ksp) infl uence the sharpness of the end point. Concentration varies over orders of magnitude, so it is useful to plot the p function: In Chapter 8, we write the p function more p function: pX 52log10[X] (7-2) correctly in terms of activity instead of concentration. For now, we use pX 5 2log[X]. where [X] is the concentration of X. Consider the titration of 25.00 mL of 0.100 0 M I2 with 0.050 00 M Ag1,

Titration reaction: I2 1 Ag1 S AgI(s) (7-3)

and suppose that we are monitoring [Ag1] with an electrode. Reaction 7-3 is the reverse of the dissolution of AgI(s), whose solubility product is rather small:

Δ 1 2 1 2 217 AgI(s) Ag 1 I Ksp 5 [Ag ][I ] 5 8.3 3 10 (7-4) 16 The equilibrium constant for the titration reaction 7-3 is largeFreeman (K 5 1/Ksp 5 1.2 3 10 ), so the equilibrium lies far to the right. Each aliquot of Ag1 reacts nearly completely with I2, leaving only a tiny amount of Ag1 in solution. At the equivalence point, there will be a sudden increase in [Ag1] because there is no I2 left to consume the added Ag1. What volume of Ag1 titrant is needed to reach the equivalence point? We calculate this 1 2 volume, designated Ve, with the fact that 1 mol of Ag reacts with 1 mol of I . 2 1 (0.025 00 L)(0.100 0 mol I /L) 5 (Ve)(0.050 00 mol Ag /L) Ve 5 volume of titrant at equivalence point ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ mol I2 mol Ag1 1 Ve 5 0.050 00 L 5 50.00 mL The titration curve has three distinct regions, depending on whether we are before, at, or Eventually, we will derive a single, uniﬁ ed W.H. equation for a spreadsheet that treats all after the equivalence point. Let’s consider each region separately. regions of the titration curve. To understand Before the Equivalence Point the chemistry, we break the curve into three 1 2 1 regions described by approximate equations Suppose that 10.00 mL of Ag have been added. There are more moles of I than Ag at this that are easy to use. point, so virtually all Ag1 is “used up” to make AgI(s). We want to fi nd the small concentration of Ag1 remaining in solution after reaction with I2. Imagine that Reaction 7-3 has gone to completion and that some AgI redissolves (Reaction 7-4). The solubility of Ag1 is determined by the concentration of free I2 remaining in the solution:

1 Ksp [Ag ] 5 2 (7-5) When V , Ve, the concentration of unreacted [I ] I2 regulates the solubility of AgI. Free I2 is overwhelmingly from the I2 that has not been precipitated by 10.00 mL of Ag1. By comparison, I2 from dissolution of AgI(s) is negligible.

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So let’s fi nd the concentration of unprecipitated I2: Moles of I2 5 original moles of I2 2 moles of Ag1 added 5 (0.025 00 L)(0.100 mol/L) 2 (0.010 00 L)(0.050 00 mol/L) 5 0.002 000 mol I2 The volume is 0.035 00 L (25.00 mL 1 10.00 mL), so the concentration is 0.002 000 mol I2 [I2] 5 5 0.057 14 M (7-6) 0.035 00 L The concentration of Ag1 in equilibrium with this much I2 is 217 Ksp 8.3 3 10 [Ag1] 5 5 5 1.4 3 10215 M (7-7) [I2] 0.057 14 5 Finally, the p function we seek is 1 1 215 pAg 52log[Ag ] 52log(1.45 3 10 ) 5 14.84 (7-8) 1 log 1.4 3 10215 5 14.84 Two signifi cant fi gures inK provide two signifi cant fi gures in [Ag ]. The two fi gures in 1 5 2 sp U ⎧ ⎨ ⎩ 1 Two signifi cant Two digits [Ag ] translate into two fi gures in the mantissa of the p function, which is written as 14.84. 2 fi gures in mantissa The preceding step-by-step calculation is a tedious way to fi nd the concentration of I . Signifi cant fi gures in logarithms were Here is a streamlined procedure that is well worth learning. Bear in mind that Ve 5 50.00 mL. discussed in Section 3-2. When 10.00 mL of Ag1 have been added, the reaction is one-fi fth complete because 10.00 mL out of the 50.00 mL of Ag1 needed for complete reaction have been added. Therefore, four- fi fths of the I2 is unreacted. If there were no dilution, [I2] would be four-fi fths of its original value. However, the original volume of 25.00 mL has been increased to 35.00 mL. If no I2 had been consumed, the concentration would be the original value of [I2] times (25.00/35.00). Accounting for both the reaction and the dilution, we can write

Original volume of I2 2 4.000 25.00 Streamlined calculation well worth using. [I ] 5 (0.100 0 M) 5 0.057 14 M a5.000b a35.00b

⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ⎧ ⎪ ⎨ ⎪ ⎩ Total volume of solution Fraction Original Dilution remaining concentration factor This is the same result found in Equation 7-6.

EXAMPLE Using the Streamlined Calculation 1 Let’s calculate pAgFreeman when VAg1 (the volume added from the buret) is 49.00 mL.

2 Solution Because Ve 5 50.00 mL, the fraction of I reacted is 49.00/50.00, and the fraction remaining is 1.00/50.00. The total volume is 25.00 1 49.00 5 74.00 mL.

1.00 25.00 [I2] 5 (0.100 0 M) 5 6.76 3 1024 M a50.00b a74.00b ⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ⎧ ⎪ ⎨ ⎪ ⎩ Fraction Original Dilution remaining concentration factor 1 2 217 24 213 [Ag ] 5 Ksp/[I ] 5 (8.3 3 10 )/(6.76 3 10 ) 5 1.23 3 10 M W.H.pAg 1 52log[Ag1] 5 12.91 The concentration of Ag1 is negligible compared with the concentration of unreacted I2, even though the titration is 98% complete.

1 TEST YOURSELF Find pAg at 49.1 mL. (Answer: 12.86)

At the Equivalence Point Now we have added exactly enough Ag1 to react with all the I2. Imagine that all AgI precipitates and some redissolves to give equal concentrations of Ag1 and I2. The value of pAg1 is found by setting [Ag1] 5 [I2] 5 x in the solubility product: 1 When V 5 Ve, [Ag ] is determined by the 1 2 solubility of pure AgI. This problem is the [Ag ][I ] 5 Ksp 2 2 1 same as if we had just added AgI(s) to water. ( x)(x) 5 8.3 3 10 17 1 x 5 9.1 3 10 9 1 pAg 52log x 5 8.04

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This value of pAg1 is independent of the original concentrations or volumes.

After the Equivalence Point Virtually all Ag1 added before the equivalence point has precipitated. The solution contains 1 all of the Ag added after the equivalence point. Suppose that VAg1 5 52.00 mL. The volume past the equivalence point is 2.00 mL. The calculation proceeds as follows:

1 1 1 Moles of Ag 5 (0.002 00 L)(0.050 00 mol Ag /L) 5 0.000 100 mol When V . Ve, [Ag ] is determined by the 1 [Ag1] 5 (0.000 100 mol)/(0.077 00 L) 5 1.30 3 1023 M 1 pAg1 5 2.89 excess Ag added from the buret. Total volume 5 77.00 mL We could justify three signifi cant fi gures for the mantissa of pAg1 because there are three signifi cant fi gures in [Ag1]. For consistency with earlier results, we retain only two fi gures. For a streamlined calculation, the concentration of Ag1 in the buret is 0.050 00 M, and 2.00 mL of titrant are being diluted to (25.00 1 52.00) 5 77.00 mL. Hence, [Ag1] is

Volume of excess Ag1 2.00 [Ag1] 5 (0.050 00 M) 5 1.30 3 1023 M a77.00b ⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎨ ⎪ ⎩ Total volume of solution Original Dilution concentration factor of Ag1

The Shape of the Titration Curve Titration curves in Figure 7-2 illustrate the effect of reactant concentration. The equivalence point is the steepest point of the curve. It is the point of maximum slope (a negative slope in this case) and is therefore an infl ection point (at which the second derivative is 0): dy Steepest slope: reaches its greatest value dx d2y Infl ection point: 5 0 dx2 In titrations involving 1 : 1 stoichiometry of reactants, the equivalence point is the steepest 1 22 S point of the titration curve. For stoichiometries other than 1 : 1, such as 2Ag 1 CrO4 Ag2CrO4(s), the curve is not symmetric. The equivalence point is not at the center of the steepest section of the curve, and it is not an infl ection point.Freeman In practice, conditions are chosen

Excess Ag+ FIGURE 7-2 Titration curves showing the Excess I– in this region in this region effect of diluting the reactants. – 0.1 M I Outer curve: 25.00 mL of 0.100 0 M I2 titrated 0.01 M I– with 0.050 00 M Ag1 14 Middle curve: 25.00 mL of 0.010 00 M I2 titrated with 0.005 000 M Ag1 2 12 0.001 M I– Inner curve: 25.00 mL of 0.001 000 M I titrated with 0.000 500 0 M Ag1 10 W.H.

+ Equivalence 8

pAg point [Ag+] = [I–]

0.001 M I– 4 0.01 M I–

2 0.1 M I–

0 10 20 30 40 50 60 70 V Ag+ (mL)

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such that titration curves are steep enough for the steepest point to be a good estimate of the equivalence point, regardless of the stoichiometry. Figure 7-3 illustrates how Ksp affects the titration of halide ions. The least soluble prod- At the equivalence point, the titration curve is uct, AgI, gives the sharpest change at the equivalence point. However, even for AgCl, the steepest for the least soluble precipitate. curve is steep enough to locate the equivalence point accurately. The larger the equilibrium constant for a titration reaction, the more pronounced will be the change in concentration near the equivalence point. − K × −17 I ( sp = 8.3 10 ) 14 EXAMPLE Calculating Concentrations During a Precipitation Titration

12 − K × −13 25.00 mL of 0.041 32 M Hg (NO ) were titrated with 0.057 89 M KIO . Br ( sp = 5.0 10 ) 2 3 2 3 21 2 S 10 Hg2 1 2IO3 Hg2(IO3)2(s) Cl− (K = 1.8 × 10−10) Iodate + sp I− 8 218 21 pAg For Hg2(IO3)2, Ksp 5 1.3 3 10 . Find [Hg2 ] in the solution after addition of (a) 34.00 mL Br− of KIO ; (b) 36.00 mL of KIO ; and (c) at the equivalence point. 6 3 3 CI−

4 Solution The volume of iodate needed to reach the equivalence point is found as follows: 2 2 2 mol IO3 1 2 2 Moles of IO3 5 21 (moles of Hg2 ) a1 mol Hg2 b 0 10 20 30 40 50 60 70 V + 1 Ag (mL) (Ve)(0.057 89 M) 5 2(25.00 mL)(0.041 32 M) Ve 5 35.69 mL ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ 2 ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ 21 ⎪ ⎪ ⎪ ⎪ ⎩ Moles of IO 3 Moles of Hg2 FIGURE 7-3 Titration curves showing the 21 effect of Ksp. Each curve is calculated for (a) When V 5 34.00 mL, the precipitation of Hg2 is not yet complete. 21 25.00 mL of 0.100 0 M halide titrated with Original volume of Hg2 0.050 00 M Ag1. Equivalence points are 35.69 2 34.00 25.00 marked by arrows. [Hg21] 5 (0.041 32 M) 5 8.29 3 1024 M 2 a 35.69 b a25.00 1 34.00b ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ Fraction Original Dilution Total volume of solution remaining concentration factor 21 of Hg2 (b) When V 5 36.00 mL, the precipitation is complete. We have gone (36.00 2 35.69) 5 2 0.31 mL past the equivalence point. The concentration of excess IO 3 is

2 Volume of excess IO 3 0.31 [IO2] 5 (0.057 89 M) 5 2.9 3 1024 M 3Freemana25.00 1 36.00b 4 ⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ Total volume of solution Original Dilution concentration factor 2 of IO 3 21 2 The concentration of Hg2 in equilibrium with solid Hg2(IO3)2 plus this much IO 3 is K 218 21 sp 1.3 3 10 211 [Hg2 ] 5 2 2 5 24 2 5 1.5 3 10 M [IO3 ] (2.94 3 10 ) 2 21 (c) At the equivalence point, there is exactly enough IO 3 to react with all Hg2 . We can imagine that all of the ions precipitate and then some Hg2(IO3)2(s) redissolves, giving two moles of iodate for each mole of mercurous ion: W.H. 21 2 Δ Hg2(IO3)2(s) Hg2 1 2IO3 x 2x 2 1 21 27 (x)(2x) 5 Ksp x 5 [Hg2 ] 5 6.9 3 10 M

21 24 TEST YOURSELF Find [Hg2 ] at 34.50 and 36.5 mL. (Answer: 5.79 3 10 M, 2.2 3 10212 M)

Our calculations presume that the only chemistry that occurs is the reaction of anion with cation to precipitate solid salt. If other reactions occur, such as complex formation or ion-pair formation, we would have to modify the calculations.

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7-4 Titration of a Mixture If a mixture of two ions is titrated, the less soluble precipitate forms fi rst. If the solubilities A liquid containing suspended particles is said are suffi ciently different, the fi rst precipitation is nearly complete before the second to be turbid because the particles scatter light. commences.

Consider the addition of AgNO3 to a solution containing KI and KCl. Because Ksp(AgI) The product with the smaller Ksp precipitates V 2 Ksp(AgCl), AgI precipitates fi rst. When precipitation of I is almost complete, the concen- ﬁ rst, if the stoichiometry of the precipitates is tration of Ag1 abruptly increases and AgCl begins to precipitate. When Cl2 is consumed, the same. Precipitation of I2 and Cl2 with Ag1 another abrupt increase in [Ag1] occurs. We expect two breaks in the titration curve, fi rst at produces two breaks in the titration curve, ﬁ rst for I2 and then for Cl2. Ve for AgI and then at Ve for AgCl. Figure 7-4 shows an experimental curve for this titration. The apparatus used to measure the curve is shown in Figure 7-5, and the theory of how this system measures Ag1 concentration is discussed in Section 15-2. The I2 end point is taken as the intersection of the steep and nearly horizontal curves at 23.85 mL shown in the inset of Figure 7-4. Precipitation of I2 is not quite complete when Cl2 begins to precipitate. (The way we know that I2 precipitation is not complete is by a calculation. That’s what these obnoxious calculations are for!) Therefore, the end of the steep portion (the intersection) is a better approximation of the equivalence point than is the middle of the steep section. The Cl2 end point is taken as the midpoint of the second steep section, at 47.41 mL. The moles of Cl2 in the sample equal the moles of Ag1 delivered between the fi rst and second end points. That is, it requires 23.85 mL of Ag1 to precipitate I2, and (47.41 2 23.85) 5 23.56 mL of Ag1 to precipitate Cl2. Comparison of the I2/Cl2 and pure I2 titration curves in Figure 7-4 shows that the I2 end point is 0.38% too high in the I2/Cl2 titration. We expect the fi rst end point at 23.76 mL, but it is observed at 23.85 mL. Two factors contribute to this high value. One is experimental error, which is always present. This discrepancy is as likely to be positive as negative. How- ever, the end point in some titrations, especially Br2/Cl2 titrations, is systematically 0 to 3% high, depending on conditions. This error is attributed to coprecipitation of AgCl with AgBr. Even though the solubility of AgCl has not been exceeded, some Cl2 becomes attached to AgBr crystallites (small crystals) as they precipitate. Each Cl2 carries down an equivalent amount of Ag1. A high concentration of nitrate reduces coprecipitation, perhaps because 2 2 NO 3 competes with Cl for binding sites on AgBr(s).

pAg+ controlled pAg+ controlled Buret − − by excess I by excess Cl in (AgNO3) in this region this regionFreeman

700 (b) pH (a) meter 600

500 Beaker 400 Glass Ag electrode electrode Solution of 300 (a) – –

mV Iodide I plus Cl end point W.H.23.76 mL 200 (23.85 mL) Stirring bar Chloride Magnetic 23.0 23.5 24.0 100 end point stirrer (47.41 mL) 0 (b) FIGURE 7-5 Apparatus for measuring the titration curves in Figure 7-4. The silver electrode responds to − 100 1 0 10 20 30 40 50 changes in Ag concentration, and the glass electrode VAg+ (mL) provides a constant reference potential in this experiment. The measured voltage changes by FIGURE 7-4 Experimental titration curves. (a) Titration curve for 40.00 mL of 0.050 2 M KI approximately 59 mV for each factor-of-10 change in 1 plus 0.050 0 M KCl titrated with 0.084 5 M AgNO3. The inset is an expanded view of the [Ag ]. All solutions, including AgNO3, were maintained region near the ﬁ rst equivalence point. (b) Titration curve for 20.00 mL of 0.100 4 M I2 at pH 2.0 by using 0.010 M sulfate buffer prepared 1 titrated with 0.084 5 M Ag . from H2SO4 and KOH.

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The second end point in Figure 7-4 corresponds to complete precipitation of both halides. 2 It is observed at the expected value of VAg1. The concentration of Cl , found from the difference between the two end points, will be slightly low in Figure 7-4, because the fi rst end point is slightly high.

7-5 Calculating Titration Curves with a Spreadsheet Now you understand the chemistry that occurs at different stages of a precipitation titration, and you should know how to calculate the shape of a titration curve. We now introduce spreadsheet calculations that are more powerful than hand calculations and less prone to error. If you are not interested in spreadsheets at this time, you can skip this section. 1 0 0 Consider the addition of VM liters of cation M (whose initial concentration is CM ) to V X 2 0 liters of solution containing anion X with a concentration C X. Ksp M1 1 X2 Δ MX(s) (7-9) Titrant Analyte 0 0 0 C M , VM C X, V X 0 1 1 The total moles of added M (5 C M ? VM) must equal the moles of M in solution (5 [M ](VM 1 0 V X) plus the moles of precipitated MX(s). (This equality is called a mass balance, even though it is really a mole balance.) In a similar manner, we can write a mass balance for X. 0 1 0 The mass balance states that the moles of an Mass balance for M: CM?VM 5 [M ](VM 1 VX) 1 mol MX(s) (7-10) element in all species in a mixture equal the ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ Total mol Moles of M Moles of M in total moles of that element delivered to the of added M in solution precipitate solution. 0 0 2 0 Mass balance for X: CX?VX 5 [X ](VM 1 VX) 1 mol MX(s) (7-11) ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ Total mol Moles of X Moles of X in of added X in solution precipitate Now equate mol MX(s) from Equation 7-10 with mol MX(s) from Equation 7-11: 0 1 0 0 0 2 0 CM?VM 2 [M ](VM 1 VX) 5 CX?VX 2 [X ](VM 1 VX) which can be rearranged to

C 0 1 [M1] 2 [X2] 2 1 5 0 X Precipitation of X with M : VM VX 0 1 2 (7-12) aCM 2 [M ] 1 [X ]b

1 1 2 0 0 A supplementary section at www.whfreeman. Equation 7-12 relates the volume of added M to [M ], [X ], and the constants V X, C X, 0 com/qca derives a spreadsheet equation for and C M. To use EquationFreeman 7-12 in a spreadsheet, enter values of pM and compute correspond- the titration of a mixture, such as that in ing values of VM, as shown in Figure 7-6 for the iodide titration of Figure 7-3. This is the Figure 7-4. reverse of the way you normally calculate a titration curve in which VM would be input and pM would be output. Column C of Figure 7-6 is calculated with the formula [M1] 5 102pM, 2 1 and column D is given by [X ] 5 Ksp/[M ]. Column E is calculated from Equation 7-12. The

FIGURE 7-6 Spreadsheet for titration of 25 A BCD E 2 1 mL of 0.1 M I with 0.05 M Ag . 1 Titration of I- with Ag+ 2 3 Ksp(Agl) = pAg [Ag+] [l-] Vm W.H.4 8.30E-17 15.08 8.32E-16 9.98E-02 0.035 5 Vo = 15 1.00E-15 8.30E-02 3.195 6 25 14 1.00E-14 8.30E-03 39.322 7 Co(l) = 12 1.00E-12 8.30E-05 49.876 8 0.1 10 1.00E-10 8.30E-07 49.999 9 Co(Ag) = 8 1.00E-08 8.30E-09 50.000 10 0.05 6 1.00E-06 8.30E-11 50.001 11 4 1.00E-04 8.30E-13 50.150 12 3 1.00E-03 8.30E-14 51.531 13 2 1.00E-02 8.30E-15 68.750 14 C4 = 10^-B4 15 D4 = $A$4/C4 16 E4 = $A$6*($A$8+C4-D4)/($A$10-C4+D4)

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fi rst input value of pM (15.08) was selected by trial and error to produce a small VM. You can start wherever you like. If your initial value of pM is before the true starting point, then VM in column E will be negative. In practice, you will want more points than we have shown so that you can plot an accurate titration curve.

7-6 End-Point Detection Precipitation titration end points are commonly found with electrodes (Figure 7-5) or indica- Titrations with Ag1 are called argentometric tors. We now describe two indicator methods for the titration of Cl2 with Ag1: titrations. Volhard titration: formation of a soluble, colored complex at the end point Fajans titration: adsorption of a colored indicator on the precipitate at the end point Because the Volhard method is a titration of Ag1, it can be adapted for the determination of Volhard Titration many anions that form insoluble silver salts. 1 The Volhard method is a titration of Ag in ,0.5 M HNO3 with standard KSCN (potassium thiocyanate). To measure Cl2, a back titration is necessary. First, Cl2 in ,0.5 M HNO is 3 (a) precipitated in a vigorously stirred solution by a known, small excess of standard AgNO3. − 1 2 S Ag 1 Cl AgCl(s) + − + − + − Vigorous stirring minimizes trapping of excess Ag1 in the precipitate. AgCl(s) is fi ltered and 1 − + − + − + washed with dilute (,0.16 M) HNO3 to collect excess Ag from the precipitate. Then Fe(NO3)3 (ferric nitrate) or Fe(NH4)(SO4)2 ? 12H2O (ferric ammonium sulfate also called − + − + − + − ferric alum) solution is added to the combined fi ltrate to give ,0.02 M Fe31. Ag1 is then titrated with standard KSCN: − + − + − + 1 2 S Ag 1 SCN AgSCN(s) + − When Ag1 has been consumed, the next drop of SCN2 reacts with Fe31 to form a red complex. Fe31 1 SCN2 S FeSCN21 Red 2 The appearance of red color is the end point. Knowing how much SCN was required for the Adsorbed 1 2 back titration tells us how much Ag was left over from the reaction with Cl . The total ions amount of Ag1 is known, so the amount consumed by Cl2 can be calculated. 31 The purpose of HNO3 in the titration solution is to prevent hydrolysis of Fe to form Fe(OH)21. Concentrated (,70 wt%) nitric acid is prepared for use by mixing with an equal (b) volume of water and boiling for a few minutes (in a hood!) to remove NO2, which has a red −+ color that would make the color change at the end point harder to see. The Volhard analysis 2 should not be conducted above room temperature to preventFreeman oxidation of SCN by warm + − + − + − HNO3. The ferric nitrate or ferric alum solutions used as indicators are stabilized by a few + − + − + − + drops of concentrated HNO3 to prevent precipitation of ferric hydroxide. In the analysis of Cl2 by the Volhard method, the end point would slowly fade if AgCl + − + − + − were not fi ltered off, because AgCl is more soluble than AgSCN. AgCl slowly dissolves and is replaced by AgSCN. To eliminate this secondary reaction, we fi lter the AgCl, and titrate − + − + − + Ag1 in the fi ltrate. An alternative to fi ltration is to add a few mL of nitrobenzene and stir vigorously to coat AgCl with nitrobenzene, which retards access to SCN2. In the analysis of + 2 2 In2− Br and I , whose silver salts are less soluble than AgSCN, it is not necessary to isolate the Adsorbed silver halide precipitate. indicator Fajans Titration W.H. FIGURE 7-7 Ions from a solution are The Fajans titration uses an adsorption indicator. To see how this works, consider the elec- adsorbed on the surface of a growing crystallite. 1 2 2 tric charge at the surface of a precipitate. When Ag is added to Cl , there is excess Cl in (a) A crystal growing in the presence of excess solution prior to the equivalence point. Some Cl2 is adsorbed onto the AgCl surface, impart- lattice anions (anions that belong in the crystal) ing a negative charge to the crystal (Figure 7-7a). After the equivalence point, there is will have a slight negative charge because excess Ag1 in solution. Adsorption of Ag1 onto the AgCl surface places positive charge on anions are predominantly adsorbed. (b) A the precipitate (Figure 7-7b). The abrupt change from negative to positive occurs at the crystal growing in the presence of excess lattice equivalence point. cations will have a slight positive charge and can therefore adsorb a negative indicator ion. Common adsorption indicators are anionic dyes that are attracted to positively charged Anions and cations in the solution that do not particles produced immediately after the equivalence point. Adsorption of the negatively belong in the crystal lattice are less likely to be charged dye onto the positively charged surface changes the color of the dye. The color adsorbed than are ions belonging to the lattice. change is the end point in the titration. Because the indicator reacts with the precipitate These diagrams omit other ions in solution. surface, we want as much surface area as possible. To attain maximum surface area, we use Overall, each solution plus its growing conditions that keep the particles as small as possible, because small particles have more crystallites must have zero total charge.

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DEMONSTRATION 7-1 Fajans Titration

The Fajans titration of Cl2 with Ag1 convincingly demonstrates Color Plate 2a shows the yellow color of the indicator in the indicator end points in precipitation titrations. Dissolve 0.5 g of NaCl solution prior to the titration. Color Plate 2b shows the milky NaCl plus 0.15 g of dextrin in 400 mL of water. The purpose of the white appearance of the AgCl suspension during titration, before dextrin is to retard coagulation of the AgCl precipitate. Add 1 mL of the end point is reached. The pink suspension in Color Plate 2c dichlorofl uorescein indicator solution containing either 1 mg/mL of appears at the end point, when the anionic indicator becomes dichlorofl uorescein in 95% aqueous ethanol or 1 mg/mL of the adsorbed on the cationic particles of precipitate. sodium salt in water. Titrate the NaCl solution with a solution containing 2 g of AgNO3 in 30 mL H2O. About 20 mL are required to reach the end point.

surface area than an equal volume of large particles. Low electrolyte concentration helps prevent coagulation of the precipitate and maintains small particle size. ؊O O O

ç œ çå œœ The most common indicator for AgCl is dichlorofl uorescein. This dye is greenish

å å yellow in solution but turns pink when adsorbed onto AgCl (Demonstration 7-1 and Color Cl œ çœ ç Cl ؊ Plate 2). The pH of the reaction must be controlled because the indicator is a weak acid å Å CO 2 2 2 and must be present in its anionic form. The dye eosin is useful in the titration of Br , I , and SCN2. It gives a sharper end point than dichlorofl uorescein and is more sensitive Dichlorofluorescein (that is, less halide can be titrated). It cannot be used for AgCl, because eosin is more strongly bound than Cl2 to AgCl. Eosin binds to AgCl crystallites even before the parti- Br Br cles become positively charged. In all argentometric titrations, but especially with adsorption indicators, strong light ؊O å O å O

ç œ çå œœ (such as daylight through a window) should be avoided. Light decomposes silver salts, and å

Br œ çœå çBr adsorbed indicators are especially light sensitive. å Å CO؊ Applications of precipitation titrations are listed in Table 7-1. Whereas the Volhard 2 method is an argentometric titration, the Fajans method has wider applications. Because the Volhard titration is carried out in acidic solution (typically 0.2 M HNO3), it avoids certain Tetrabromofluorescein (eosin) 22 22 32 interferences that affect other titrations. Silver salts of CO3 , C2O4 , and AsO4 are soluble in acidic solution, so these anions do not interfere. Freeman TABLE 7-1 Applications of precipitation titrations Species analyzed Notes Volhard Method Br2, I2, SCN2, CNO2, Precipitate removal is unnecessary. 32 AsO4 , 2 32 2 22 Cl , PO4 , CN , C2O4 , Precipitate removal required. 22 22 22 CO3 , S , CrO4 2 1 2 BH4 Back titration of Ag left after reaction with BH4 : 2 1 1 1 2 S 1 2 1 W.H. BH4 8Ag 8OH 8Ag(s) H2BO3 5H2O Fajans Method Cl2, Br2, I2, SCN2, Titration with Ag1. Detection with dyes such as fl uorescein, 42 Fe(CN)6 dichlorofl uorescein, eosin, bromophenol blue. 21 Zn Titration with K4Fe(CN)6 to produce K2Zn3[Fe(CN)6]2. End-point detection with diphenylamine. 22 SO4 Titration with Ba(OH)2 in 50 vol% aqueous methanol using alizarin red S as indicator. 21 Hg2 Titration with NaCl to produce Hg2Cl2. End point detected with bromophenol blue. 32 22 PO4 , C2O4 Titration with Pb(CH3CO2)2 to give Pb3(PO4)2 or PbC2O4. 32 End point detected with dibromofl uorescein (PO4 ) or 22 fl uorescein (C2O4 ).

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Terms to Understand adsorption indicator equivalence point standardization trace analysis argentometric titration Fajans titration standard solution Volhard titration back titration gravimetric titration titrant volumetric analysis blank titration indicator titration direct titration primary standard titration curve end point reagent-grade chemical titration error

Summary The volume of reagent (titrant) required for stoichiometric reaction point, there is excess analyte. Its concentration is the product (frac- of analyte is measured in volumetric analysis. The stoichiometric tion remaining) 3 (original concentration) 3 (dilution factor). The point of the reaction is called the equivalence point. What we mea- concentration of titrant can be found from the solubility product of sure by an abrupt change in a physical property (such as the color of the precipitate and the known concentration of excess analyte. At an indicator or the potential of an electrode) is the end point. The the equivalence point, concentrations of both reactants are governed difference between the end point and the equivalence point is a by the solubility product. After the equivalence point, the concen- titration error. This error can be reduced by subtracting results of a tration of analyte can be determined from the known concentration blank titration, in which the same procedure is carried out in the of excess titrant and the solubility product. absence of analyte, or by standardizing the titrant, using the same The end points of two common argentometric titrations of reaction and a similar volume as that used for analyte. anions that precipitate with with Ag1 are marked by color changes. The validity of an analytical result depends on knowing the In the Volhard titration, excess standard AgNO3 is added to the amount of a primary standard. A solution with an approximately anion and the resulting precipitate is fi ltered off. Excess Ag1 in the desired concentration can be standardized by titrating a primary fi ltrate is back titrated with standard KSCN in the presence of Fe31. standard. In a direct titration, titrant is added to analyte until the When Ag1 has been consumed, SCN2 reacts with Fe31 to form a reaction is complete. In a back titration, a known excess of reagent red complex. The Fajans titration uses an adsorption indicator to is added to analyte, and the excess is titrated with a second standard fi nd the end point of a direct titration of anion with standard reagent. Calculations of volumetric analysis relate the known moles AgNO3. The indicator color changes right after the equivalence of titrant to the unknown moles of analyte. point when the charged indicator is adsorbed onto the oppositely Concentrations of reactants and products during a precipitation charged surface of the precipitate. titration are calculated in three regions. Before the equivalence

Exercises 2 7-A. Ascorbic acid (vitamin C) reacts with I 3 according to the equa- (c) A vitamin C tablet containing ascorbic acid plus inert binder tion was ground to a powder, and 0.424 2 g was titrated by 31.63 mL of 2 FreemanI 3. Find the weight percent of ascorbic acid in the tablet. OH 7-B. A solution of NaOH was standardized by gravimetric titration HO O of a known quantity of the primary standard, potassium hydrogen • O phthalate: I3 H2O CO3H CO2Na HO OH ß ß ß aO —B ß Ascorbic acid OH 2 CO2K CO2K C6H8O6 O O 3I 2H O OH Potassium hydrogen phthalate C H O K, FM 204.22 W.H.HO OH 8 5 4 Dehydroascorbic acid The NaOH was then used to fi nd the concentration of an unknown C H O 6 8 7 solution of H2SO4: S Starch is used as an indicator in the reaction. The end point is H2SO4 1 2NaOH Na2SO4 1 2H2O marked by the appearance of a deep blue starch-iodine complex (a) Verify from the structure of potassium hydrogen phthalate that 2 when the fi rst fraction of a drop of unreacted I 3 remains in the its formula is C8H5O4K. solution. (b) Titration of 0.824 g of potassium hydrogen phthalate required (a) Verify that the structures above have the chemical formulas 38.314 g of NaOH solution to reach the end point detected by written beneath them. You must be able to locate every atom in the phenolphthalein indicator. Find the concentration of NaOH (mol formula. Use atomic masses from the periodic table on the inside NaOH/kg solution).

cover of this book to fi nd the formula mass of ascorbic acid. (c) A 10.00-mL aliquot of H2SO4 solution required 57.911 g of 2 (b) If 29.41 mL of I 3 solution are required to react with 0.197 0 g of NaOH solution to reach the phenolphthalein end point. Find the 2 pure ascorbic acid, what is the molarity of the I 3 solution? molarity of H2SO4.

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7-C. A solid sample weighing 0.237 6 g contained only malonic 7-E. Construct a graph of pAg1 versus milliliters of Ag1 for the acid and aniline hydrochloride. It required 34.02 mL of 0.087 71 M titration of 40.00 mL of solution containing 0.050 00 M Br2 and 2 1 NaOH to neutralize the sample. Find the weight percent of each 0.050 00 M Cl . The titrant is 0.084 54 M AgNO3. Calculate pAg component in the solid mixture. The reactions are at the following volumes: 2.00, 10.00, 22.00, 23.00, 24.00, 30.00, 2 S 2 40.00 mL, second equivalence point, 50.00 mL. CH2(CO2H)2 1 2OH CH2(CO2 )2 1 2H2O 2 Malonic acid FM 104.06 Malonate 7-F. Consider the titration of 50.00 (60.05) mL of a mixture of I 2 1 and SCN with 0.068 3 (60.000 1) M Ag . The fi rst equivalence —S NH –H O Cl NH3 Cl O 2 2 point is observed at 12.6 (60.4) mL, and the second occurs at 27.7 Aniline hydrochloride Aniline (60.3) mL. FM 129.59 (a) Find the molarity and the uncertainty in molarity of thiocyanate in the original mixture. 7-D. A 50.0-mL sample of 0.080 0 M KSCN is titrated with 0.040 0 M Cu1. The solubility product of CuSCN is 4.8 3 10215. (b) Suppose that the uncertainties are all the same, except that the At each of the following volumes of titrant, calculate pCu1, and uncertainty of the fi rst equivalence point (12.6 6 ? mL) is variable. 1 1 What is the maximum uncertainty (milliliters) of the fi rst equiva- construct a graph of pCu versus milliliters of Cu added: 0.10, 2 10.0, 25.0, 50.0, 75.0, 95.0, 99.0, 100.0, 100.1, 101.0, 110.0 mL. lence point if the uncertainty in SCN molarity is to be #4.0%?

Problems Volumetric Procedures and Calculations The excess acid required 39.96 mL of 0.100 4 M NaOH for complete titration to a phenolphthalein end point. Find the weight 7-1. Explain the following statement: “The validity of an analytical percent of calcite in the limestone. result ultimately depends on knowing the composition of some primary standard.” 7-12. Arsenic(III) oxide (As2O3) is available in pure form and is a useful (but carcinogenic) primary standard for oxidizing agents 7-2. Distinguish between the terms equivalence point and end point. 2 such as MnO 4 . The As2O3 is dissolved in base and then titrated with 2 2 7-3. How does a blank titration reduce titration error? MnO 4 in acidic solution. A small amount of iodide (I ) or iodate (IO2) is used to catalyze the reaction between H AsO and MnO2. 7-4. What is the difference between a direct titration and a back 3 3 3 4 2 Δ 22 titration? As2O3 1 4OH 2HAsO3 1 H2O 22 1 Δ 7-5. What is the difference between a reagent-grade chemical and a HAsO3 1 2H H3AsO3 2 1 S 21 primary standard? 5H3AsO3 1 2MnO4 1 6H 5H3AsO4 1 2Mn 1 3H2O

7-6. Why are ultrapure acid solvents required to dissolve samples (a) A 3.214-g aliquot of KMnO4 (FM 158.034) was dissolved in for trace analysis? 1.000 L of water, heated to cause any reactions with impurities to occur, cooled, and fi ltered. What is the theoretical molarity of this 7-7. How many milliliters of 0.100 M KI are needed to react with 40.0 2 21 2 S solution if no MnO 4 was consumed by impurities? mL of 0.040 0 M Hg2(NO3)2 if the reaction is Hg2 1 2I Hg2I2(s)? Freeman (b) What mass of As2O3 (FM 197.84) would be just suffi cient to 7-8. For Reaction 7-1, how many milliliters of 0.165 0 M KMnO4 react with 25.00 mL of the KMnO4 solution in part (a)? are needed to react with 108.0 mL of 0.165 0 M oxalic acid? How (c) It was found that 0.146 8 g of As2O3 required 29.98 mL of KMnO4 many milliliters of 0.165 0 M oxalic acid are required to react with 2 solution for the faint color of unreacted MnO4 to appear. In a blank 108.0 mL of 0.165 0 M KMnO ? 2 4 titration, 0.03 mL of MnO4 was required to produce enough color to 2 be seen. Calculate the molarity of the permanganate solution. 7-9. Ammonia reacts with hypobromite, OBr , by the reaction 2NH3 1 2 S 2 3OBr N2 1 3Br 1 3H2O. What is the molarity of a hypobromite 7-13. A 0.238 6-g sample contained only NaCl and KBr. It was solution if 1.00 mL of the OBr2 solution reacts with 1.69 mg of NH ? 3 dissolved in water and required 48.40 mL of 0.048 37 M AgNO3 7-10. Sulfamic acid is a primary standard that can be used to stan- for complete titration of both halides [giving AgCl(s) and dardize NaOH. W.H. AgBr(s)]. Calculate the weight percent of Br in the solid sample. 1 2 2 2 7-14. A solid mixture weighing 0.054 85 g contained only ferrous H3NSO3 1 OH ¡ H2NSO3 1 H2O Sulfamic acid ammonium sulfate and ferrous chloride. The sample was dissolved FM 97.094 21 41 in 1 M H2SO4, and the Fe required 13.39 mL of 0.012 34 M Ce 31 41 21 S 31 31 What is the molarity of a sodium hydroxide solution if 34.26 mL for complete oxidation to Fe (Ce 1 Fe Ce 1 Fe ). react with 0.333 7 g of sulfamic acid? Calculate the weight percent of Cl in the original sample. 7-11. Limestone consists mainly of the mineral calcite, CaCO . The FeSO4 ? (NH4)2SO4 ? 6H2O FeCl2 ? 6H2O 3 Ferrous ammonium sulfate Ferrous chloride carbonate content of 0.541 3 g of powdered limestone was FM 392.13 FM 234.84 measured by suspending the powder in water, adding 10.00 mL of 1.396 M HCl, and heating to dissolve the solid and expel CO : 7-15. A cyanide solution with a volume of 12.73 mL was treated 2 with 25.00 mL of Ni21 solution (containing excess Ni21) to convert 1 21 c CaCO3(s) 1 2H ¡ Ca 1 CO2 1 H2O the cyanide into tetracyanonickelate(II): Calcium carbonate 2 1 2 FM 100.087 4CN 1 Ni2 S Ni CN 2 1 24

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The excess Ni21 was then titrated with 10.15 mL of 0.013 07 M 7-20. In precipitation titrations of halides by Ag1, the ion pair ethylenediaminetetraacetic acid (EDTA): AgX(aq) (X 5 Cl, Br, I) is in equilibrium with the precipitate. Use Appendix J to fi nd the concentrations of AgCl(aq), AgBr(aq), and 21 1 42 S 22 Ni EDTA Ni(EDTA) AgI(aq) during the precipitations. Ni(CN)22 does not react with EDTA. If 39.35 mL of EDTA were 4 7-21. Sulfate in soil on Mars. A barium sulfate precipitation titration required to react with 30.10 mL of the original Ni21 solution, cal- 2 described at the opening of this chapter is shown in the fi gure. The culate the molarity of CN in the 12.73-mL cyanide sample. 2 initial concentration of Cl before adding BaCl2 was 0.000 19 M in 7-16. Managing a salt-water aquarium. A tank at the New Jersey 25 mL of aqueous extract of Martian soil. At the end point, when State Aquarium has a volume of 2.9 million liters.5 Bacteria are there is a sudden rise in Ba21, [Cl2] 5 0.009 6 M. used to remove nitrate that would otherwise build up to toxic (a) Write the titration reaction. levels. Aquarium water is fi rst pumped into a 2 700-L deaeration (b) How many mmol of BaCl2 were required to reach the end point? tank containing bacteria that consume O in the presence of 2 2 (c) How many mmol of SO2 were contained in the 25 mL? added methanol: 4 (d) If SO 22 is derived from 1.0 g of soil, what is the wt% of SO22 Bacteria 4 4 2CH3OH 1 3O2 ¡ 2CO2 1 4H2O (1) in the soil? Methanol −1.0 Leaching solution added − Anoxic (deoxygenated) water from the deaeration tank fl ows into a Cl calibration − Cl 1 500-L denitrifi cation reactor containing colonies of Pseudomonas solution added −2.0 bacteria in a porous medium. Methanol is injected continuously and Soil added and BaCl2 begins to nitrate is converted into nitrite and then into nitrogen: − enter cell Cl = 0.009 6 M 2 Bacteria 2 −3.0 at end point 3NO3 1 CH3OH ¡ 3NO2 1 CO2 1 2H2O (2) − Nitrate nitrite Cl = 0.000 19 M + before BaCl addition Ba2 Bacteria 2 2 2 log (concentration, M) −4.0 2NO2 1 CH3OH ¡ N2 1 CO2 1 H2O 1 2OH (3)

Deaeration End point CH3OH −5.0 tank 0:00 2:00 4:00 6:00 8:00 10:00 12:00 14:00 16:00 Aquarium Time (h) Denitrification Barium sulfate precipitation titration from Phoenix Mars Lander. [Data from tank CH3OH Reference 1. Courtesy S. Kounaves, Tufts University.] Titration of a Mixture (a) Deaeration can be thought of as a slow, bacteria-mediated titration of O2 by CH3OH. The concentration of O2 in seawater at 248C 7-22. Describe the chemistry that occurs in each of the following is 220 mM. How many liters of CH3OH (FM 32.04, density 5 regions in curve (a) in Figure 7-4: (i) before the fi rst equivalence 0.791 g/mL) are required by Reaction 1 for 2.9 million Freemanliters of point; (ii) at the fi rst equivalence point; (iii) between the fi rst and aquarium water? second equivalence points; (iv) at the second equivalence point; and (b) Write the net reaction showing nitrate plus methanol going to (v) past the second equivalence point. For each region except (ii), 1 nitrogen. How many liters of CH3OH are required by the net write the equation that you would use to calculate [Ag ]. reaction for 2.9 million liters of aquarium water with a nitrate 7-23. The text claims that precipitation of I2 is not complete before m 2 concentration of 8 100 M? Cl begins to precipitate in the titration in Figure 7-4. Calculate the (c) In addition to consuming methanol for Reactions 1 through 3, concentration of Ag1 at the equivalence point in the titration of I2 the bacteria require 30% more methanol for their own growth. What alone. Show that this concentration of Ag1 will precipitate Cl2. is the total volume of methanol required to denitrify 2.9 million 6 liters of aquarium water? 7-24. A procedure for determining halogens in organic compounds uses an argentometric titration. To 50 mL of anhydrous ether is added Shape of a PrecipitationW.H. Curve a carefully weighed sample (10–100 mg) of unknown, plus 2 mL of sodium dispersion and 1 mL of methanol. (Sodium dispersion is fi nely 7-17. Describe the chemistry that occurs in each of the following divided solid sodium suspended in oil. With methanol, it makes 2 1 regions in Figure 7-2: (i) before the equivalence point; (ii) at the sodium methoxide, CH3O Na , which attacks the organic compound, equivalence point; and (iii) past the equivalence point. Write the liberating halides.) Excess sodium is destroyed by slow addition of 1 equation to fi nd [Ag ] in each region. 2-propanol, after which 100 mL of water are added. (Sodium should not be treated directly with water, because the H produced can 7-18. Consider the titration of 25.00 mL of 0.082 30 M KI with 2 1 explode in the presence of O : 2Na 1 2H O S 2NaOH 1 H .) This 0.051 10 M AgNO . Calculate pAg at the following volumes of 2 2 2 3 procedure gives a two-phase mixture, with an ether layer fl oating on added AgNO : (a) 39.00 mL; (b) V (c) 44.30 mL. 3 e; top of the aqueous layer that contains the halide salts. The aqueous 1 7-19. A 25.00-mL solution containing 0.031 10 M Na2C2O4 was layer is adjusted to pH 4 and titrated with Ag , using the electrodes in titrated with 0.025 70 M Ca(NO3)2 to precipitate calcium oxalate: Figure 7-5. How much 0.025 70 M AgNO3 solution will be required to 21 22 S 21 Ca 1 C2O4 CaC2O4(s). Find pCa at the following volumes reach each equivalence point when 82.67 mg of 1-bromo-4-chlorobutane of Ca(NO3)2: (a) 10.00, (b) Ve, (c) 35.00 mL. (BrCH2CH2CH2CH2Cl; FM 171.46) are analyzed?

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7-25. Calculate pAg1 at the following points in titration (a) in 7-30. Use the equation in Problem 7-29 to calculate the titration 22 1 Figure 7-4: (a) 10.00 mL; (b) 20.00 mL; (c) 30.00 mL; (d) second curve for 10.0 mL of 0.100 M CrO4 titrated with 0.100 M Ag to equivalence point; (e) 50.00 mL. produce Ag2CrO4(s). 7-26. A mixture having a volume of 10.00 mL and containing 1 21 End-Point Detection 0.100 0 M Ag and 0.100 0 M Hg2 was titrated with 0.100 0 M KCN to precipitate Hg2(CN)2 and AgCN. 7-31. Why does the surface charge of a precipitate change sign at (a) Calculate pCN2 at each of the following volumes of added the equivalence point? KCN: 5.00, 10.00, 15.00, 19.90, 20.10, 25.00, 30.00, 35.00 mL. 7-32. Examine the procedure in Table 7-1 for the Fajans titration of (b) Should any AgCN be precipitated at 19.90 mL? Zn21. Do you expect the charge on the precipitate to be positive or negative after the equivalence point? Using Spreadsheets 7-33. Describe how to analyze a solution of NaI by using the Vol- 7-27. Derive an expression analogous to Equation 7-12 for the titra- hard titration. 1 0 0 2 tion of M (concentration 5 CM , volume 5 VM ) with X (titrant 0 7-34. A 30.00-mL solution of HBr was treated with 5 mL of freshly concentration 5 CX ). Your equation should allow you to compute 2 boiled and cooled 8 M HNO3, and then with 50.00 mL of 0.365 0 M the volume of titrant (VX) as a function of [X ]. AgNO3 with vigorous stirring. Then 1 mL of saturated ferric alum 7-28. Use Equation 7-12 to reproduce the curves in Figure 7-3. was added and the solution was titrated with 0.287 0 M KSCN. Plot your results on a single graph. When 3.60 mL had been added, the solution turned red. What was x2 m1 the concentration of HBr in the original solution? How many milli- 7-29. Consider precipitation of X with M : grams of Br2 were in the original solution? xMm1 1 mXx2 Δ M X (s) K 5 [Mm1]x[Xx2]m x m sp 7-35. What is wrong with this procedure? According to Table 7-1, Write mass balance equations for M and X and derive the equation carbonate can be measured by a Volhard titration. Removal of the precipitate is required. To analyze an unknown solution of Na2CO3, xC 0 1 m[Mm1] 2 x[Xx2] 5 0 X I acidifi ed the solution with freshly boiled and cooled HNO3 to VM VX 0 m1 x2 amCM 2 m[M ] 1 x[X ]b give ,0.5 M HNO3. Then I added excess standard AgNO3, but no Ag2CO3 precipitate formed. What happened? x2 m1 x 1/m where [X ] 5 (Ksp/[M ] ) . Freeman

W.H.

160 CHAPTER 7 Let the Titrations Begin